Problem: Real numbers $x$ and $y$ have an arithmetic mean of 18 and a geometric mean of $\sqrt{92}$. Find $x^2+y^2$.
The givens tell us that $\frac{x+y}{2}=18$ and $\sqrt{xy}=\sqrt{92}$, or $x+y=36$ and $xy=92$. $(x+y)^2=x^2+2xy+y^2$, so  \[
x^2+y^2=(x+y)^2-2xy=36^2-2\cdot92=1296-184=\boxed{1112}
\]